The points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of ∆ ABC. Find the coordinates of the point P on AD such that AP : PD = 2 : 1
Solution:
Given, the vertices of the triangle ABC are A(x₁, y₁), B (x₂, y₂) and C (x₃, y₃).
The point P lies on AD such that AP : PD = 2 : 1
We have to find the coordinates of the point P.
Let the coordinates of the point P be (x, y)
The coordinates of the point P which divides the line segment joining the points A (x₁ , y₁) and B (x₂ , y₂) internally in the ratio m₁ : m₂ are
[(m₁x₂ + m₂x₁)/m₁ + m₂ , (m₁y₂ + m₂y₁)/m₁ + m₂]
The point P(x, y) divides the line segment A(x₁, y₁) and D((x₂ + x₃)/2, (y₂ + y₃)/2) in the ratio 2 : 1
Here, m₁ : m₂ = 2 : 1.
So, (x, y) = [2(x₂ + x₃)/2 + 1(x₁)/2 + 1, 2(y₂ + y₃)/2 + 1(y₁)/2 + 1]
(x, y) = [(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3]
Therefore, the coordinates of the point P are (x₁ + x₂ + x₃)/3 and (y₁ + y₂ + y₃)/3.
✦ Try This: A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC. D(7/2, 9/2) is the midpoint of BC. Find the coordinates of the point P on AD such that AP : PD = 2 : 1
Given, the vertices of the triangle ABC are A(4, 2), B (6, 5) and C (1, 4).
Midpoint of BC, D = (7/2, 9/2)
The point P lies on AD such that AP : PD = 2 : 1
We have to find the coordinates of the point P.
Let the coordinates of the point P be (x, y)
The coordinates of the point P which divides the line segment joining the points A (x₁ , y₁) and B (x₂ , y₂) internally in the ratio m₁ : m₂ are
[(m₁x₂ + m₂x₁)/m₁ + m₂ , (m₁y₂ + m₂y₁)/m₁ + m₂]
The point P(x, y) divides the line segment A(4, 2) and D(7/2, 9/2) in the ratio 2 :1
Here, m₁ : m₂ = 2 : 1.
So, (x, y) = [(2(3.5) + 1(4))/2 + 1, (2(4.5) + 1(2))/2 + 1]
(x, y) = [(7 + 4)/3, (9 + 2)/3]
(x, y) = [11/3, 11/3]
Therefore, the coordinates of the point P are 11/3 and 11/3.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7
NCERT Exemplar Class 10 Maths Exercise 7.4 Problem 3(ii)
The points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of ∆ ABC. Find the coordinates of the point P on AD such that AP : PD = 2 : 1
Summary:
The points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of ∆ ABC. The coordinates of the point P on AD such that AP : PD = 2 : 1 are (x₁+x₂+x₃)/3 and (y₁+y₂+y₃)/3
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